Answer:
[tex]v=\sqrt{k\frac{e^2}{m_e r}}[/tex], [tex]2.18\cdot 10^6 m/s[/tex]
Explanation:
The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:
[tex]k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}[/tex]
where
k is the Coulomb constant
e is the magnitude of the charge of the electron
e is the magnitude of the charge of the proton in the nucleus
r is the distance between the electron and the nucleus
v is the speed of the electron
[tex]m_e[/tex] is the mass of the electron
Solving for v, we find
[tex]v=\sqrt{k\frac{e^2}{m_e r}}[/tex]
Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately
[tex]r=5.3\cdot 10^{-11}m[/tex]
while the electron mass is
[tex]m_e = 9.11\cdot 10^{-31}kg[/tex]
and the charge is
[tex]e=1.6\cdot 10^{-19} C[/tex]
Substituting into the formula, we find
[tex]v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s[/tex]